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\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(\frac{15}{34}+\frac{1}{3}+\frac{19}{34}+\left(-\frac{4}{3}\right)+\frac{3}{7}\)
\(\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{1}{3}-\frac{4}{3}\right)+\frac{3}{7}\)
\(1+\left(-1\right)+\frac{3}{7}\)
\(=0+\frac{3}{7}=\frac{3}{7}\)
\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=\)\(\frac{15}{34}+\frac{1}{3}+\frac{19}{34}+\left(-\frac{4}{3}\right)+\frac{3}{7}\)
\(=\)\(\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{1}{3}-\frac{4}{3}\right)+\frac{3}{7}\)
\(=\)\(1+\left(-1\right)+\frac{3}{7}\)
\(=\)\(0+\frac{3}{7}\)
\(=\)\(\frac{3}{7}\)
Câu 1:
\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)
\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)
Câu 2:
\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)
a, \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{20}{15}+\dfrac{3}{7}\)
= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{3}{7}\right)-\dfrac{20}{15}\)
= 1 + \(\dfrac{16}{21}-\dfrac{20}{15}\)
= \(\dfrac{3}{7}\)
b, \(\dfrac{27}{25}+\dfrac{4}{21}-\dfrac{2}{25}+\dfrac{17}{21}-\dfrac{1}{2}\)
= \(\left[\dfrac{27}{25}+\left(-\dfrac{2}{25}\right)\right]+\left(\dfrac{4}{21}-\dfrac{7}{21}\right)-\dfrac{1}{2}\)
= 1 + \(\left(\dfrac{-1}{7}\right)-\dfrac{1}{2}\) = \(\dfrac{5}{14}\)
Answer:
\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\) (Mình đã sửa đề nhé.)
\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=1+\left(-1\right)+\frac{3}{7}\)
\(=\frac{3}{7}\)
\(\frac{3}{8}.27\frac{1}{5}-51\frac{1}{5}.38+19\)
\(=\frac{3}{8}\left(-24\right).38+19\)
\(=-342+19\)
\(=-323\)
\(-\frac{2}{7}.\frac{21}{8}.-\frac{7}{2}\)
\(=\left(\frac{-2}{7}.\frac{-7}{2}\right)\frac{21}{8}\)
\(=\frac{21}{8}\)
\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{32}{17}+\dfrac{2}{3}\)
\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\dfrac{7}{21}+\dfrac{-32}{17}+\dfrac{2}{3}\)
\(=1+\dfrac{1}{3}+\dfrac{-32}{17}+\dfrac{2}{3}\)
\(=1+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{-32}{17}\)
\(=1+1+\dfrac{-32}{17}\)
\(2+\dfrac{-32}{17}=\dfrac{2}{17}\)